Solution of 1785D/1786F (a typical FFT problem)

Revision en13, by YocyCraft, 2023-02-07 13:51:13

Problem link:1785D - Wooden Spoon

First, WLOG we can assume we arrange the tournament in such way: In every matches, the left player wins. We can achieve this by doing the following operation to the binary tree of the tournament: for each non-leaf node of the binary tree, if the winner of this node is its right child, we "flip" this node by swapping its left subtree and right subtree. Since fliping doesn't change the winner of each match, this will not change the "wooden spoon", and after this operation, the "wooden spoon" is the right-most node. Since there are $$$2^{n}-1$$$ nodes in the binary tree, we can merge $$$2^{2^n-1}$$$ situations into one by this operation.

For example, we can do operation to this tree:

_________1

_____3_______1

-__5___3___1___2

__7_5_3_6_1_8_4_2

-->

_________1

_____1_______3

-__1___2___3___5

__1_8_2_4_3_6_5_7

Where $$$1$$$ (the left-most node) is the champion, and $$$7$$$ (the right-most node) is the "wooden spoon".

Then we assume the right-most node is k, and there's $$$dp[n][k]$$$ different arrangements (after operation). If $$$k$$$ is the $$$j$$$-th smallest element in the right half of the tree, then we have $$$\sum_{i=1}^{2^{n-1}}dp[n-1][i]$$$ ways to arrange the left half, and $$$dp[n-1][j]$$$ ways to arrange the right half (since $$$k$$$ is also the right-most node in the right-subtree). But in how many ways we can distribute $$$2^{n}$$$ elements into $$$2$$$ halves? Well, since $$$1$$$ is the left-most element, there are $$$k-2$$$ elements we could put in the right part (which are in the range $$$[2, k-1]$$$), and there are actually $$$j-1$$$ elements of them in the right part, so we have $$$\binom{k-2}{j-1}$$$ ways to choose them. Similarly, we have $$$\binom{2^{n}-k}{2^{n-1}-j}$$$ ways to choose elements from $$$[k+1, 2^k]$$$. Therefore, we can get such formula:

$$$dp[n][k]=\displaystyle \sum_{j=1}^{2^{n-1}}(\displaystyle \sum_{i=1}^{2^{n-1}}dp[n-1][i]) \cdot dp[n-1][j] \cdot \binom{k-2}{j-1} \cdot \binom{2^{n}-k}{2^{n-1}-j}$$$

Then we can calculate them by FFT. The answer is $$$2^{2^{n}-1} \cdot dp[n][k]$$$.

Code example

History

 
 
 
 
Revisions
 
 
  Rev. Lang. By When Δ Comment
en13 English YocyCraft 2023-02-07 13:51:13 28
en12 English YocyCraft 2023-02-07 13:49:21 1 Tiny change: '{2^{n-1}}(sum_{i=1}^' -> '{2^{n-1}}(\sum_{i=1}^'
en11 English YocyCraft 2023-02-07 13:48:50 210 Tiny change: 'can merge 2^(2^n-1) situation' -> 'can merge $$2^{2^n-1} situation'
en10 English YocyCraft 2023-02-07 12:48:19 20
en9 English YocyCraft 2023-02-07 08:15:54 6
en8 English YocyCraft 2023-02-07 08:14:03 6 Tiny change: 'h way: In the every match, the left' -> 'h way: In every matches, the left'
en7 English YocyCraft 2023-02-07 08:08:59 29 Tiny change: 'ht child, swap its left ' -> 'ht child, we "flip" this node by swapping its left '
en6 English YocyCraft 2023-02-07 08:05:49 6715
en5 English YocyCraft 2023-02-07 08:04:17 28
en4 English YocyCraft 2023-02-07 08:03:43 4
en3 English YocyCraft 2023-02-07 08:03:24 64
en2 English YocyCraft 2023-02-07 08:02:39 76
en1 English YocyCraft 2023-02-07 08:01:52 1884 Initial revision (published)